Question

A displacement vector, at an angle of 30∘ with y-axis has a x-component of 10 units. Then the magnitude of the vector is?
A. 5.0
B. 10
C. 11.5
D. 20

Answer 1

Hint:Displacement vectors have both magnitude and direction. In the figure displacement vector is represented graphically with an arrow which gives the direction. To find the magnitude of the displacement vector we only find the length of this vector. To get the magnitude of the displacement vector in two dimensions is represented as the resultant vector. We can simply measure the magnitude or length of the displacement vector by using the Pythagorean theorem.

Formula used:
Using trigonometric Ratios
\[\cos \theta = \dfrac{\text{Base}}{\text{Perpendicular}}\]
Where \[\theta \] is an angle between base and perpendicular.

Complete step by step solution:
x-component = 10 units
Displacement vector making angle to y-axis =\[{30^0}\]
Also, displacement vector making angle to x-axis
As \[\Delta \] PPQ is a right angled triangle,
\[\begin{array}{l}\angle POQ = {90^0} - {30^0}{\rm{ }} = {60^0}\end{array}\]
Let the displacement vector be \[\vec r\] making an angle \[\theta \] with the x-axis.

In right angled triangle POQ, we get
\[\cos {60^0} = \dfrac{{10}}{r}\]
As we know \[\cos {60^0} = \dfrac{1}{2}\]
By substituting this value,
\[\dfrac{1}{2} = \dfrac{{10}}{r}\]
\[\therefore r = 20\]
Therefore the magnitude of the displacement vector, r is 20 units.

Hence option D is the correct answer.

Note: The displacement vector from one point to another is defined as an arrow with its tail at the first point and its tip at the second. The magnitude or the length of the displacement vector is defined as the distance between the points. It is represented by the length of the arrow. The direction of the arrow gives the direction of the displacement vector.